LA 4328 Priest John's Busiest Day (Greedy)

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2329

　　很好的一道贪心题。在贪心的时候，因为时间安排是要连续的，所以用区间的中点作为排序的依据。中点靠前的区间要先贪心，标记当前最后的时间点。

#define REP(i, n) for (int i = 0; i < (n); i++)

struct Cerem {
    int begin, end;
    int len, mid;
    Cerem(int b = 0, int e = 0) {
        begin = b, end = e;
        len = (e - b >> 1) + 1;
        mid = b + len;
    }
} cerem[N];

bool cmp(Cerem a, Cerem b) {
    return a.mid < b.mid;
}

void input(int n) {
    int b, e;
    REP(i, n) {
        scanf("%d%d", &b, &e);
        cerem[i] = Cerem(b, e);
    }
    sort(cerem, cerem + n, cmp);
}

bool work(int n) {
    int end = 0;
    REP(i, n) {
        if (end <= cerem[i].begin) end = cerem[i].begin + cerem[i].len;
        else if (end + cerem[i].len <= cerem[i].end) end += cerem[i].len;
        else return false;
    }
    return true;
}

int main() {
    int n;
    while (~scanf("%d", &n) && n) {
        input(n);
        work(n) ? puts("YES") : puts("NO");
    }
    return 0;
}

——written by Lyon